Consider the delta well potential energy:

$\mathit{U}\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}\mathbf{\{}\begin{array}{ll}\mathbf{0}& \mathbf{x}\mathbf{\ne }\mathbf{0}\\ \mathbf{-}\mathbf{\infty }& \mathbf{x}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}\end{array}$

Although not completely realistic, this potential energy is often a convenient approximation to a verystrong, verynarrow attractive potential energy well. It has only one allowed bound-state wave function, and because the top of the well is defined as U = 0, the corresponding bound-state energy is negative. Call its value -E₀.

(a) Applying the usual arguments and required continuity conditions (need it be smooth?), show that the wave function is given by

$\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}\mathbf{=}{\mathbf{\left(}\frac{\mathbf{2}\mathbf{m}{\mathbf{E}}_{\mathbf{0}}}{{\mathbf{h}}^{\mathbf{2}}}\mathbf{\right)}}^{\mathbf{1}\mathbf{/}\mathbf{4}}{\mathit{e}}^{\mathbf{-}\mathbf{(}\sqrt{\mathbf{2}\mathbf{m}{\mathbf{E}}_{\mathbf{0}}}\mathbf{/}\mathbf{\hslash }\mathbf{)}\mathbf{\left|}\mathbf{x}\mathbf{\right|}}$

(b) Sketch $\mathit{\psi }\mathbf{\left(}\mathbf{x}\mathbf{\right)}$and U(x) on the same diagram. Does this wave function exhibit the expected behavior in the classically forbidden region?

There is a delta well potential of the form

$\mathrm{U}\left(\mathrm{x}\right)=\left\{\begin{array}{l}\begin{array}{ll}0& \mathrm{x}\ne 0\end{array}\\ \begin{array}{ll}-\mathrm{\infty }& \mathrm{x}=0\end{array}\end{array}\right.$....................................(I)

The top of the well is defined as U = 0 and the corresponding bound-state energy is negative (-E₀).

The wave function of a particle of mass m and energy E outside a finite potential well of height U₀ is

$\psi \left(x\right)=\left\{\begin{array}{ll}A{e}^{\frac{\sqrt{2m\left(U_0-E\right)}}{\hslash }x}& x<0\\ B{e}^{-\frac{\sqrt{2m\left(U_0-E\right)}}{\hslash }x}& x>0\end{array}\right.$ .....(II)

Here is the reduced Planck's constant.

For the delta potential, $U_0$= 0 and E =$-E$and the wave function in equation (II) reduces to

localid="1660047013782" $\psi \left(x\right)=\left\{\begin{array}{ll}A{e}^{\frac{\sqrt{2m\left(U_0-E\right)}}{\hslash }x}& x<0\\ B{e}^{-\frac{\sqrt{2m\left(U_0-E\right)}}{\hslash }x}& x>0\end{array}\right.$

The function has to be continuous at x = 0 and thus

A = B

The function thus becomes

localid="1660047017169" $\psi \left(x\right)=A{e}^{-\frac{\sqrt{2m{E}_0}}{\hslash }\left|x\right|}$

Normalize this to get

localid="1660047021009" $\begin{array}{rcl}1& =& \underset{-\infty }{\overset{\infty }{\int }}{\left(A{e}^{-\frac{\sqrt{2m{E}_0}}{\hslash }\left|x\right|}\right)}^{2}dx\\ & =& 2A^2\underset{0}{\overset{\infty }{\int }}{e}^{-\frac{2\sqrt{2m{E}_0}}{\hslash }x}dx\\ & & \end{array}$

Let

localid="1660047024841" $\begin{array}{rcl}\frac{2\sqrt{2m{E}_0}}{\hslash }x& =& z\\ dx& =& \frac{\hslash }{2\sqrt{2m{E}_0}}dz\\ & & \end{array}$

Thus

localid="1660047032303" $\begin{array}{rcl}1& =& 2A^2\frac{\hslash }{2\sqrt{2m{E}_0}}\underset{0}{\overset{\infty }{\int }}{e}^{-z}dz\\ & =& \frac{A^2\hslash }{\sqrt{2m{E}_0}}\\ A& =& {\left(\frac{\sqrt{2m{E}_0}}{\hslash }\right)}^{1/2}\\ & & \end{array}$

The final wave function is

localid="1660047039951" $\psi \left(x\right)={\left(\frac{\sqrt{2m{E}_0}}{\hslash }\right)}^{1/2}{e}^{-\frac{\sqrt{2m{E}_0}}{\hslash }\left|x\right|}$

The wave function obtained above is plotted as follows

The wave function exponentially falls off in the classically forbidden region as expected.